Find the final Celsius temperature of the aluminum and water

In summary, Q_total=-602.549J/K in the system of particles consisting of all the particles in the aluminum plus all the particles in the water.
  • #1
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Homework Statement



A 2.3kg aluminum pan at 135° is plunged into 3.0kg of water at 15°. The specific heat of aluminum is 900 J/(kg K).



Part A
Find the final Celsius temperature of the aluminum and water.
T_f = °Celsius

Part B
Find the entropy change of the aluminum pan.
ΔS_Al = J/K

Part C
Find the entropy change of the water.
ΔS_H20 = J/K

Part D
Find the entropy change in the system of particles consisting of all the particles in the aluminum plus all the particles in the water.
ΔS = J/K





Homework Equations



Q=mCΔT
ΔS=Q/T



The Attempt at a Solution




So my first couple attempts was that we use the equation Q=mC(T_f-T_i)

other related equations i know is DeltaS=Q/T but, how can i find the temperature final of both Water and aluminum.

I tried setting it up like this

m_water*C_water(T_f-T-i) + m_aluminum*C_aluminum(T_f-T_i) = Q_total

but i don't know how to get anything done from there.

Please help me with the first part and i should be able to most of it done on my own.
 

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  • #2
I found the answer to the first two. And the first two will eventually help you with the rest.

attached in this form is a file showing the integration for part b.

Part A
Q_water+Qaluminum=0
m_waterC_water(T_f-T_water)+m_aluminumC_aluminum(T_f-T_aluminum)=0
using distribution from algebra and algebraically solving for T_F,
T_f=(m_waterC_waterT_water+m_aluminumCaluminumT_aluminum)/(m_waterC_water+m_aluminumC_aluminum)
you get Tf = 31.96 degrees Celsius.

Where, if your values were,
m_aluminum=2.3kg
T_aluminum=135 degrees Celsius
C_aluminum=900J/kg*k
m_water=3kg
T_water=15 degrees celsius
C_water=4190J/kg*kPart B


=(2.3kg)(900J/kg*k)ln((31.96+273)/(135+273))=-602.549J/K

Part C
use the same format, as b, but with different mass and specific heat, and T_i values, T_f is the same.

Part D is the sum of part B and C.
 

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Last edited:
  • #3
This all looks good.
 

1. What is the formula for finding the final Celsius temperature of aluminum and water?

The formula for finding the final Celsius temperature of aluminum and water is: Tfinal = (m1 x c1 x T1 + m2 x c2 x T2) / (m1 x c1 + m2 x c2), where m1 is the mass of aluminum, c1 is the specific heat capacity of aluminum, T1 is the initial temperature of aluminum, m2 is the mass of water, c2 is the specific heat capacity of water, and T2 is the initial temperature of water.

2. What is the specific heat capacity of aluminum and water?

The specific heat capacity of aluminum is 0.902 J/g°C and the specific heat capacity of water is 4.184 J/g°C. These values represent the amount of energy required to raise the temperature of 1 gram of the substance by 1 degree Celsius.

3. How do I determine the initial temperatures and masses of aluminum and water?

The initial temperatures and masses of aluminum and water can be measured using a thermometer and a scale. The initial temperature of each substance can be recorded before they are mixed together in a container. The masses can also be measured before mixing or by subtracting the mass of the container from the total mass of the mixture.

4. What is the significance of finding the final Celsius temperature of aluminum and water?

Finding the final Celsius temperature of aluminum and water is important because it allows us to understand how the two substances interact when they are mixed together. It also helps us determine the amount of energy transferred between the two substances, which can be useful in various scientific experiments and applications.

5. What are some factors that may affect the final Celsius temperature of aluminum and water?

The final Celsius temperature of aluminum and water can be affected by several factors, including the initial temperatures and masses of the two substances, the specific heat capacities of aluminum and water, and any heat transfer or loss that may occur during the mixing process. Other factors such as pressure and external temperature can also play a role in the final temperature.

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